In this paper (A simple provably secure key exchange by Ding et al.) At page number 8, the author gives correctness of the technique as follows
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then SK A = SKB with overwhelming probability i.e. if Alice and Bob run the protocol honestly, then they will share an identical key.

The above equation uses Lemma 1 which is as follows

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How does the author deduce the above equation using Lemma 1. This equation gives correctness of the technique. Can anyone please help.

  • $\begingroup$ There's a reference to Lemma 1 in what you've reproduced. That lemma's the key. $\endgroup$ – Michael Snook May 25 '17 at 11:54
  • $\begingroup$ Thanks for replying. How he used lemma 1 to produce this equation $\endgroup$ – vivek May 25 '17 at 12:11
  • $\begingroup$ I don't have time for a full answer right now. You should add lemma 1 to the question so someone else can use it in their answer. $\endgroup$ – Michael Snook May 25 '17 at 12:22
  • $\begingroup$ Please see the related posted question on crypto.stackexchange.com/questions/48147/… Thanks $\endgroup$ – vivek Jun 10 '17 at 6:11
  • $\begingroup$ Please see this bquestion crypto.stackexchange.com/questions/48566/… $\endgroup$ – vivek Jun 23 '17 at 11:35

First: Lemma 1 says that $||\mathbf{x}|| \leq \alpha q \sqrt{n}$ with overwhelming probability if $\mathbf{x}$ is drawn from the discrete Gaussian since $\frac{1}{2^n}$ is negligible.

Next, from properties of absolute values, $|a + b| \leq |a| + |b|$. So, leaving the $2$ out for now and writing $\mathbf{s_A}^T\mathbf{e_B}$ as $\mathbf{s_A}\cdot\mathbf{e_B}$:

$|\mathbf{s_A}\cdot\mathbf{e_B} + e'_A + \mathbf{s_B}\cdot\mathbf{e_A} + e'_B| \leq |\mathbf{s_A}\cdot\mathbf{e_B}| + |e'_A| + |\mathbf{s_B}\cdot\mathbf{e_A}| + |e'_B|$.

Now, for Euclidean norms, Cauchy-Schwarz says $|\mathbf{a\cdot b}| \leq |\mathbf{a}|\cdot |\mathbf{b}|$, so we have, for example, $|\mathbf{s_A}\cdot\mathbf{e_B}| \leq |\mathbf{s_A}| \cdot |\mathbf{e_B}| \leq (\alpha q \sqrt{n})\cdot (\alpha q \sqrt{n})$, the last inequality coming from Lemma 1.

Let's tackle $e'_A$ and $e'_B$. I could sample a vector $\mathbf{e'}$ from $\mathcal{D_{\mathbb{Z^n},\alpha q}}$ and Lemma 1 would apply to it; if $e'_A$ is a member of $\mathbf{e'}$, it is certainly smaller than $||\mathbf{e'}||$:

$|e'_A| \leq ||\mathbf{e'_A}|| \leq \alpha q \sqrt{n} \leq (\alpha q \sqrt{n})\cdot (\alpha q \sqrt{n})$. Same for $e'_B$.

Thus I have four terms, all $\leq (\alpha q \sqrt{n})\cdot (\alpha q \sqrt{n})$. Multiply back in that $2$ and you have the result.


They do a similar procedure later on in Section 4, and explicitly write out norms, for future reference.

  • $\begingroup$ Sir I understand that value of |SA* eB|≤|SA| |eB|≤(αq√n). (αq√n) as ||SA||≤(αq√n) and ||eB||≤(αq√n) but how would alone |e'B|≤(αq√n). (αq√n). Thanks $\endgroup$ – vivek May 26 '17 at 4:28
  • $\begingroup$ I was making a transitive argument: if $A \leq B$ and $B \leq C$, then $A \leq C$. Since $|e'_B| \leq ||\mathbb{e'_B}||$ (true since $e'_B$ is a member of $\mathbb{e'_B}$, and $\mathbb{e'_B} \leq \alpha q \sqrt{n}$ (from Lemma 1, and $\alpha q \sqrt{n} \leq \alpha q \sqrt{n} \cdot \alpha q \sqrt{n}$ (true for numbers bigger than 1), we have $|e'_B| \leq \alpha q \sqrt{n} \cdot \alpha q \sqrt{n}$. Does that make sense? It's a chain of inequalities. $\endgroup$ – user47922 May 26 '17 at 4:42
  • $\begingroup$ Also: only $e'_B$ is in the paper, and is a scalar. $\mathbb{e'_B}$ is a vector I introduced so that I could use Lemma 1; recall that it deals with vector samples. In the paper, they drew the scalar $e'_B$ directly from $\mathcal{D_{\mathbb{Z},\alpha q}}$. Instead, I drew a vector $\mathbb{e'_B}$ from $\mathcal{D_{\mathbb{Z^n},\alpha q}}$ to use Lemma 1. $\endgroup$ – user47922 May 26 '17 at 4:46
  • $\begingroup$ ok sir understood. Thanks for putting up lot of effort. $\endgroup$ – vivek May 26 '17 at 6:21
  • $\begingroup$ I have confusion in lemma 2 on page number 6. Can you please explain its proof. $\endgroup$ – vivek Jun 9 '17 at 8:59

This lemma is used to conclude that a sample from $\mathcal{D}_{\mathbb{Z}^n,\alpha q}$ is (with overwhelming probability) less than or equal to $\alpha q \sqrt{n}$.

Now, because all values $\textbf{s}_{\textbf{A}},\textbf{s}_{\textbf{B}},\textbf{e}'_{\textbf{A}},\textbf{e}'_{\textbf{B}}$ are sampled from $\mathcal{D}_{\mathbb{Z}^n,\alpha q}$ , so $\textbf{s}_{\textbf{A}}^T\textbf{e}_{\textbf{B}}$ is less than or equal to $(\alpha q \sqrt{n})(\alpha q \sqrt{n})$. This bound is also true for each of other values $\textbf{e}_{\textbf{A}}^T\textbf{s}_{\textbf{B}}, \textbf{e}'_{\textbf{A}},\textbf{e}'_{\textbf{B}}$ and so the upper bound is determined.

  • $\begingroup$ Oops! Didn't see your answer. I spent about an hour trying to figure out how to type $\mathcal{D_{\mathbb{Z^n},\alpha q}}$. $\endgroup$ – user47922 May 25 '17 at 21:42
  • $\begingroup$ @hamidreza I accept galvatron explanation as answer, as it contains detailed explanation. Thanks for answering. $\endgroup$ – vivek May 26 '17 at 6:23
  • $\begingroup$ @hamidreza Please see the related posted question on crypto.stackexchange.com/questions/48147/…. Thanks $\endgroup$ – vivek Jun 10 '17 at 6:10

A question came to when reading this is that: in section 4 (lemma 9), they claim that $8n\beta^2\leq q/4-2$, then $SK_a=SK_b$.

My question is that since they use $l_\infty$ norm as they said at beginning of section 4, does $||a\cdot b||_\infty \leq ||a||_\infty\cdot ||b||_\infty$ holds? It seems that for infinity norm, above inequality may not hold. If so, how do they compute $8n\beta^2\leq q/4-2$?

  • $\begingroup$ Answers are not the place to ask questions. You should post this as a new question with a link to the current question. $\endgroup$ – mikeazo Aug 22 '17 at 11:45

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